QAQ 贪心与动态规划

"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.

Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).

Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.

Input

The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.

Output

Print a single integer — the number of subsequences "QAQ" in the string.

Sample 1

QAQAQYSYIOIWIN

4

Sample 2

QAQQQZZYNOIWIN

3

Note

In the first example there are 4 subsequences "QAQ":

"QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN",

"QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".

思路：

找QAQ，简化说就是找A，然后左右Q的个数相乘就可以了。

主要是要简化问题，找到一个比较合理的解决方案。

#include<iostream>
#include<string.h>
using namespace std;
const int N=100;
int n;
char a[N+5];
int main()
{
	memset(a,0,sizeof(a));
	cin>>a;
	int l=0,r=0,ans=0;
	for(int i=0;a[i]!='\0';i++)
	{
		l=0,r=0;
		if(a[i]=='A')
		{
			for(int j=i-1;j>=0;j--)
			{
				if(a[j]=='Q')
				{
					l++;
				}
			}
			for(int j=i+1;a[j]!='\0';j++)
			{
				if(a[j]=='Q')
				{
					r++; 
				}
			}
			//printf("%d %d %d\n",i,l,r);
			ans+=l*r;
		}
	}
	printf("%d\n",ans);
	return 0;
}